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Given that the $\Delta G_{\mathrm{f}}^{\circ}$ for $\mathrm{Pb}^{2+}(a q)$ and $\mathrm{Cl}^{-}(a q)$ is $-24.3 \mathrm{kJ} / \mathrm{mole}$ and $-131.2 \mathrm{kJ} / \mathrm{mole}$ respectively, determine the solubility product, $K_{\mathrm{sp}},$ for $\mathrm{PbCl}_{2}(s)$

$1.5 \times 10^{-5}$

Chemistry 102

Chapter 16

Thermodynamics

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So for this question, were asked to find the soy ability constant of this reaction we hear. And so Oh, no, that the Delta G, um, for standard for PB two pluses. Negative 24.3 and negative 131.2 1st yell minus so and that's all getting weather problem. So next we can look at the form of that we need to use. And it's gonna be this where we essentially defined Kay because Kay is going to be equal to K. S. P in this case would use a liability constant. So we know are already and we know till race would just fill those in here and we're looking to find ke sp. But we're missing water. One other thing in that country. So to find the change in Gibbs free energy of formulation for the reaction on the stair used for use, this information we're given and one more value that we confined in appendix A G for a PB seal too. So then our, um, math is going to look like this. Who we have negative 24.3 close two times. Nega 1 31.2 and subtract that from negative 314.2. That's a value get from appendix cheap. You're gonna end up with 27.5 killer Jules Fergie. We want this to be in jewels, so we're just gonna move the decimal replaces over and, um, filling our equation. So I just moved to a new slide here. But using that formula, we're gonna get 27.5 times 10 to the third equals negative are with 8.314 times T, which is to 98 bucks. It's room temperature, and then we have Ellen K. We're trying Sulfur Island, Kate. And when we do that, we actually get K S P equal to a 1.5 times 10 to the negative fifth.

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